In dogs, albino white coat color is an autosomal recessive trait, while black coat is a dominant trait. Homozygous recessive mutations in at least two different genes have been reported to cause white coat color in poodles. A woman owned a purebred albino female poodle. The woman wanted to have more white poodle puppies. She took her dog to a breeder, who promised her to mate the woman's purebred albino female poodle with a purebred albino stud male. When six puppies were born, all of them were black. The woman sued the breeder, claiming that the breeder failed to breed her poodle with a purebred albino stud male but with a black dog, giving her six unwanted black puppies. You are called in as an expert witness. Based on your knowledge in genetics, which of the following explanations for the case would you consider to be the most reasonable to offer to the court? When two purebred homozygous recessive dogs mate, all puppies should show the recessive trait. Since all of the six puppies has the black coat color, the breeder is for sure guilty.

In dogs, albino white coat color is an autosomal recessive trait, while black coat is a dominant trait. Homozygous recessive mutations in at least two different genes have been reported to cause white coat color in poodles. A woman owned a purebred albino female poodle. The woman wanted to have more white poodle puppies. She took her dog to a breeder, who promised her to mate the woman's purebred albino female poodle with a purebred albino stud male. When six puppies were born, all of them were black. The woman sued the breeder, claiming that the breeder failed to breed her poodle with a purebred albino stud male but with a black dog, giving her six unwanted black puppies. You are called in as an expert witness. Based on your knowledge in genetics, which of the following explanations for the case would you consider to be the most reasonable to offer to the court? When two purebred homozygous recessive dogs mate, all puppies should show the recessive trait. Since all of the six puppies has the black coat color, the breeder is for sure guilty.

Name: 1. The court should request genetic tests in the woman's female albin
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Description: 1. The court should request genetic tests in the woman's female albino dog and the breeder's albino stud. In any case if the white coat allele in the two dogs are in different genes, the breeder should be innocent.2. To determine if the breeder is i

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 1CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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For numbers 34-50. Given the following pedigree, answer the following questions. 1st generation would be the grandparents, 2" generation would be the children of the grandparents and the 3" generation would be the children of the 2" generation. Trait that would be track is autosomal dominant and recessive. Cleft chin (CC or Ce) = Dominant No cleft chin (cc) = Recessive II II 10 42-45. Who are the persons failed to inherit cleft chin from the grand mother? A. Number 5, 1, 8,9 and 7 B. Number 2, 9, 7, 10 and 3 C. Number 2, 5, 7, and 9 D. Number 1, 3 and 10 E. Number 2 and 5
B. Labrador retrievers have 3 varieties of fur colour: yellow, chocolate or black. Two genes are involved, each with a dominant (Y and B) and a recessive allele (y and b). The allele Y codes for the ability to produce hair pigment: Yy and YY dogs have dark pigmented fur but all yx, individuals are lighter coloured and yellow. The B allele codes for the ability to make hair with the darkest colouration: BB and Bb individuals have black fur, bb dogs have a lighter shade which is chocolate. In the blank Punnet Square below, calculate and state the ratio of phenotypes that occur with a cross between two individuals with the genotype XyBb
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of a color-blind father. Her mother had normal color vision, but her maternal grandfather was color-blind. What is the probability that your friend is color-blind? (a) 1 (b) (c) (d) (e) 0
re ||| E 6. Label the genotypes for this pedigree of an X-linked recessive disorder (red- green colorblindness). (a) 2 2 3 1 3 a. How do you know? b. Label the genotypes. 4 O To 2 4 5 5 6 6 7. Is the following pedigree autosomal recessive, autosomal dominant or X-linked recessive? 2 8 O T 58
Background Reminders: Dominant allele will express itself in the phenotype even when in combination with a different allele (recessive). Example: PP and Pp (purple flowers) Recessive allele does not express itself in the phenotype when combined with another allele (dominant) Example: pp (white flower) -> expressed only when recessive is by itself Problem 1- Autosomal Dominant: Widow’s peak (hairline) Having widow’s peak is a dominant trait (use W and w to represent alleles) A man who has a widow’s peak has a child with a woman that does not have widow’s peak. Their child does not have widow’s peak. What is the genotype of the father? _______________ B.) A man with widow’s peak (WW) has a child with a woman who does not have a widow’s peak. What is the probability their child will have a widow’s peak? ____________ Fill out this Punnett square: C.) If both parents are heterozygous for widow’s peak (Ww)… What is the probability…
File Home Insert Layout Search I U ... ... ^dへ 3. The American Curl is a breed of cat that has curly ears. The allele for curly ears is dominant (E) and the allele for non-curly ears is recessive (e). Suppose a male cat with the heterozygous genotype is bred to a female cat with the homozygous recessive genotype. Predict the number of kittens that will have the phenotype for curly ears. In your answer: • define the word phenotype • write the genotypes for both parents; draw a Punnett square that shows a cross between the two cats (1) • write the phenotypic ratio of the offspring. If the cross produces six kittens, predict how many of them will have the curly ear 100% + Give Feedback to Microse Page 1 of 1 234 words English (U.S.) Text Predictions: On 2:2 12/1 re to search DELL Home Er F11 Priser F12 Insert F8 F9 F10 FZ F3 F4 F5 F2 & >
In humans, the ability to roll your tongue is an autosomal dominant trait (T); the inability to rol your tongue is recessive (tt), having a cleft chin ai an autosomal dominant (N) and the lack of the cleft chinn is recessive (nn). A man can roll his tongue and has a clift chin; his father can not roll his tongue and didn't have a cleft chin. The man is married to a woman who can roll her tongue and who has a cleft chin; her mother coud not roll her tongue and didn't have a cleft chin. What are the possible genotuypes and phenotypes of the children? Woman's genotypes? Man's genotypes? Woman's gamete alleles? Man's gamete alleles?
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In humans, Cystic Fibrosis an autosomal recessive trait, currently with medicine advances people affected by this disease can live until the 40s and 50s even. A couple in which the mother is a known carrier for this disease plans to have 4 children with a man who is not affected either but had his father who had the disease. A) Please state the genotypes of the parents. Use C to denote the dominant allele and c as the recessive B) Determine the probability that the first 2 children born have Cystic Fibrosis and the 2 last children are not affected. C) In general, what proportion of the children are expected to be female carriers of Cystic Fibrosis trait?